Refer to the diagram shown below.
Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.
Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.
Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s
100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s
V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°
Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s
1st question is False, 2nd question is True
This causes reverse faults<span>, which are the reverse of </span>normal faults<span>, because in this case, the hanging wall slides upward relative to the footwall. Shear </span>stress<span> is when rock slabs slide past each other horizontally. There is no vertical movement of either the hanging wall or footwall, and we get a strike-slip </span>fault<span>.</span>